This question tests graph-derivative reasoning, particularly interpreting the sign of the derivative to determine the function's monotonicity and extrema. Given f'>0 for x<0, f'(0)=0, and f'<0 for x>0, f is increasing left of 0 up to a local maximum at x=0, then decreasing right of 0. This sign pattern around x=0 confirms a change from positive to negative derivative, hallmark of a local maximum. The behavior holds on the closed interval [-3,3] without additional critical points implied. A tempting distractor like choice B fails because it describes a local minimum, which would require f' to change from negative to positive at x=0, opposite the given signs. To check sketches transferably, use a sign chart for f' to map out increasing and decreasing intervals and identify extremum types at zeros.

This problem combines monotonicity and concavity information to determine the derivative's graph. Since f decreases on (-2, 0) and increases on (0, 5), we know f'(x) < 0 for -2 < x < 0 and f'(x) > 0 for 0 < x < 5, with f'(0) = 0. Additionally, f is concave up for x > 0, meaning f'' > 0 for x > 0, which tells us f' is increasing for x > 0. An increasing line crossing the x-axis at x = 0 satisfies all conditions: it's negative before x = 0, positive after x = 0, and always increasing. Choice B (decreasing line) would make f concave down everywhere, contradicting the given concavity. To find f' from combined information, first mark where f' = 0 from extrema, then use monotonicity for f's sign pattern and concavity for whether f' increases or decreases.

This question tests graph-derivative reasoning by exploring how f' behaves around a horizontal tangent while maintaining monotonicity in f. A horizontal tangent at x = 0 means f'(0) = 0, and f increasing on both sides requires f' ≥ 0 nearby with no sign change. Thus, f' can be positive for x ≠ 0 near 0 and zero at 0, allowing f to increase through the point without an extremum. This is possible even if f' approaches 0 from positive values on both sides. A tempting distractor like choice A fails because changing from positive to negative would create a local maximum in f, contradicting the increasing on both sides. To check sketches, confirm that no sign change in f' around a zero maintains monotonicity in f, while verifying tangent conditions.

This question tests graph-derivative reasoning by relating critical points and monotonicity intervals of f to the signs and zeros of f'. Critical points occur where f' = 0, and decreasing intervals of f correspond to f' < 0, while increasing intervals mean f' > 0. The decrease on (-1,3) requires f' < 0 there, increases elsewhere require f' > 0 outside, and critical points at -1 and 3 imply f' zeros there with sign changes. This results in f' positive for x < -1, negative on (-1,3), and positive for x > 3, matching the monotonicity. A tempting distractor like choice A fails because it has f' positive on (-1,3), which would make f increasing there instead of decreasing. To check sketches, ensure zeros of f' align with critical points of f and that the signs of f' match the increasing or decreasing behavior in the specified intervals.

This question tests graph-derivative reasoning by linking concavity changes in f to monotonicity shifts in f'. Concave up on (0,2) means f'' > 0, so f' is increasing there, and concave down on (2,5) means f'' < 0, so f' is decreasing there. The change at x = 2 implies a local maximum in f' at 2, consistent with increasing before and decreasing after. The condition f'(2) = 3 specifies a positive value at that point, which fits without requiring a zero. A tempting distractor like choice A fails because decreasing on (0,2) would imply f'' < 0 and concave down, opposite to the given concave up. To check sketches, align increasing segments of f' with concave up regions of f and decreasing segments with concave down, noting points of change for potential extrema in f'.

This question tests graph-derivative reasoning by connecting concavity of f to the monotonicity of f'. Concave up regions in f correspond to f'' > 0, meaning f' is increasing there, while concave down regions mean f'' < 0, so f' is decreasing. For f concave up on (-1,0), f' increases on that interval, and concave down on (0,3), f' decreases there, with the change at x = 0 implying a local extremum in f' at 0. The condition f'(1) = 0 adds a zero in f' within the concave down interval, which is consistent as long as the overall monotonicity of f' aligns with the concavity. A tempting distractor like choice A fails because it has f' decreasing on (-1,0), which would imply f'' < 0 and concave down, opposite to the given concave up. To check sketches, match increasing intervals of f' with concave up regions of f and decreasing intervals of f' with concave down regions, verifying any specified values or points.

This question requires understanding how the second derivative affects the first derivative's graph. Since f''(x) represents the derivative of f'(x), when f''(x) > 0 for x < 0, the function f' is increasing on that interval. Similarly, when f''(x) < 0 for x > 0, the function f' is decreasing on that interval. At x = 0, where f''(0) = 0 and f'' changes from positive to negative, f' has a local maximum because it transitions from increasing to decreasing. Choice B might seem plausible if you confuse the roles of f' and f'', but remember that f'' tells us about the monotonicity of f', not f itself. When sketching derivatives, always remember that the second derivative controls whether the first derivative is increasing or decreasing.

This question tests graph-derivative reasoning by connecting features of f' to concavity and inflection points in f. Since f' is differentiable, its local maximum at x = 2 with f'(2) = 0 means f' increases before 2 and decreases after, implying f'' > 0 left and f'' < 0 right. This sign change in f'' at x = 2 indicates an inflection point in f where concavity shifts from up to down. The zero in f' at 2 marks a critical point in f, but without sign change information, it does not guarantee a max or min. A tempting distractor like choice A fails because a local maximum in f would require f' to change from positive to negative at 2, which is not necessarily true given only a local max in f'. To check sketches, associate extrema in f' with inflection points in f and verify if sign changes in f' imply extrema in f.

This problem requires translating intervals of increase/decrease into the derivative's sign. When f is increasing, f' > 0, and when f is decreasing, f' < 0. Since f increases on (-∞, -1), decreases on (-1, 3), then increases on (3, ∞), we need f' to be positive before x = -1, negative between x = -1 and x = 3, and positive after x = 3. This means f' must have zeros at x = -1 and x = 3, crossing from positive to negative at x = -1 and from negative to positive at x = 3. A downward-opening parabola with zeros at these points satisfies this pattern: it's positive between its roots and negative outside them. Choice A (upward-opening parabola) would give the opposite sign pattern, being negative between its roots. To verify derivative graphs from monotonicity, check that the derivative is positive where the function increases and negative where it decreases.

This question tests graph-derivative reasoning by analyzing how zeros and signs of f' affect the monotonicity and extrema of f. A zero in f' indicates a critical point in f, but without a sign change, there is no local extremum, and consistent positive f' means f is increasing overall. Here, f'(x) = 0 only at x = 1 but f' > 0 elsewhere ensures f is increasing everywhere, with the zero at 1 not creating an extremum due to no sign change. This describes f as strictly increasing with a horizontal tangent at 1 but no local max or min. A tempting distractor like choice B fails because a local minimum would require f' to change from negative to positive at 1, contradicting the given always positive f' except at the zero. To check sketches, examine the sign of f' around its zeros to determine if f has extrema or continues monotonically, confirming overall behavior.